uva-270【数学】

Lining Up

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

"How am I ever going to solve this problem?" said the pilot. 

Indeed, the pilot was not facing an easy task. She had to drop packages at specific points scattered in a dangerous area. Furthermore, the pilot could only fly over the area once in a straight line, and she had to fly over as many points as possible. All points were given by means of integer coordinates in a two-dimensional space. The pilot wanted to know the largest number of points from the given set that all lie on one line. Can you write a program that calculates this number? 


Your program has to be efficient! 

Input

Input consist several case,First line of the each case is an integer N ( 1 < N < 700 ),then follow N pairs of integers. Each pair of integers is separated by one blank and ended by a new-line character. The input ended by N=0.

Output

output one integer for each input case ,representing the largest number of points that all lie on one line.

Sample Input

51 12 23 39 1010 110

Sample Output

3

大意：求坐标点共线的最大数量

思路：任意找两个点作为一条边，然后判断其他点是否在在这条直线上

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int n;int x[710],y[710];bool judge(int x1,int y1,int x2,int y2,int x3,int  y3){if((y1-y3)*(x2-x3)==(y2-y3)*(x1-x3))return 1;return 0;}int main(){while(scanf("%d",&n),n){for(int i=1;i<=n;i++)scanf("%d%d",&x[i],&y[i]);int ans=0;for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){int cnt=0;for(int k=1;k<=n;k++){if(judge(x[i],y[i],x[j],y[j],x[k],y[k]))cnt++;}ans=max(ans,cnt);}}printf("%d\n",ans);}return 0; }